# Question 37fe4

Sep 21, 2017

Here's what I would try.

#### Explanation:

Let's say that the sample that is too big to fit in the graduated cylinder has a volume $V$ $\text{mL}$. After you weight the sample, you find that it has a mass $M$ $\text{g}$.

By definition, the density of the mineral, $\rho$, is equal to the ratio that exists between the mass of any sample of this mineral and the volume it occupies.

So for the initial sample, you have--I'll skip the units for simplicity

rho = M/V" "" "color(darkorange)("(*)")

Now, cut a piece of this sample that is small enough to fit in a graduated cylinder. Place this small piece on a scale and record its mass, let's say $m$ $\text{g}$.

Use the graduated cylinder to determine its volume.

Let's say that this small piece has a volume $v$ $\text{mL}$.

Since the density of the mineral must be the same regardless of the mass of the sample and the volume it occupies, you can say that--I'll skip the units for simplicity

$\rho = \frac{m}{v}$

You can now use equation $\textcolor{\mathrm{da} r k \mathmr{and} a n \ge}{\text{(*)}}$ to say that

$\frac{M}{V} = \frac{m}{v}$

Rearrange to find $V$, the volume of the initial sample

M/V = m/v implies V = (Mcolor(red)(cancel(color(black)("g"))))/(mcolor(red)(cancel(color(black)("g")))) * vcolor(white)(.)"mL"#

$V = \left(\frac{M}{m} \cdot v\right) \textcolor{w h i t e}{.} \text{mL}$

And there you have it, the volume of the initial sample as a function of its mass $M$ and the mass $m$ and volume $v$ of a smaller piece of this mineral.